What is the integral of 7/((8-x^2)^{3/2)} ?

The integral of 7/((8-x^2)^{3/2)} is (7x)/(8(8-x^2)^{1/2)}+C

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integral of 7/((8-x^2)^{3/2)}

\int \frac{7}{( 8 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{7 x}{8 ( 8 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{7}{( 8 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int \frac{1}{( 8 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{8 cos ^{2} ( u )} d u

= 7 \cdot \int \frac{1}{8 cos ^{2} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \frac{1}{8} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= 7 \cdot \frac{1}{8} tan (u)

Substitute back

u = arcsin (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)

= 7 \cdot \frac{1}{8} tan (arcsin (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x))

Simplify

7 \cdot \frac{1}{8} tan (arcsin (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) : \frac{7 x}{8 ( 8 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{7 x}{8 ( 8 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{7 x}{8 ( 8 - x ^{2} ) ^{\frac{1}{2}}} + C

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What is the integral of 7/((8-x^2)^{3/2)} ?
The integral of 7/((8-x^2)^{3/2)} is (7x)/(8(8-x^2)^{1/2)}+C